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 March 3rd, 2014, 11:41 AM #1 Newbie   Joined: Mar 2014 Posts: 6 Thanks: 0 limit involving 2 sequences Hello! I have encountered the following problem involving some limits: Let $(a_n)(n=>=1)},(b_n)(n>=1),a_0 = 1,b_0 = 1, a_{(n+1)} = a_n+b_n, b_{(n+1)} = (n^2 + n +1) a_n+b_n,n>=1.$The problems asks this limit: $\lim_{n\to\infty}n \sqrt[n]{\frac{a_1 a_2 ...a_n}{b_1 b_2 ...b_n} }$ Well I have tried several ways like Cesaro Stolz but it didn't work. After that I've tried to use a matrix:Let A be $\left( \begin{array}{ccc} 1 & n^2+n+1 \\ n^2+n+1 & 1\\ \end{array} \right)/extract_itex]Then $A^n\ \left( \begin{array}{ccc} a_0\\ b_0 \\ \end{array} \right)\ = \ \left( \begin{array}{ccc} a_{(n+1)} \\ b_{(n+1)}\\ \end{array} \right)\$ But I couldn't solve the problem... Any help would be appreciated!  March 3rd, 2014, 04:33 PM #2 Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,821 Thanks: 1047 Math Focus: Elementary mathematics and beyond Re: limit involving 2 sequences $\lim_{n\to\infty}\,n\,\cdot\,$$\frac{1}{n!}$$^{1/n}\,=\,\lim_{n\to\infty}\,\exp$$\ln(n)\,+\,\frac1n \ln\(\frac{1}{n!}$$\)\,=\,e^1\,=\,e$ If you can do an series expansion at n = ? of $\ln(n)\,+\,\frac1n\ln$$\frac{1}{n!}$$$ you've got most of a proof. March 3rd, 2014, 09:08 PM #3 Newbie Joined: Mar 2014 Posts: 6 Thanks: 0 Re: limit involving 2 sequences Quote:  Originally Posted by greg1313 $\lim_{n\to\infty}\,n\,\cdot\,$$\frac{1}{n!}$$^{1/n}\,=\,\lim_{n\to\infty}\,\exp$$\ln(n)\,+\,\frac1n \ln\(\frac{1}{n!}$$\)\,=\,e^1\,=\,e$ If you can do an series expansion at n = ? of $\ln(n)\,+\,\frac1n\ln$$\frac{1}{n!}$$$ you've got most of a proof. Unfortunately I haven't reached this level yet. I don't know how to do a series expansion and I was thinking there might be some other way to solve this exercise.  March 5th, 2014, 09:41 AM #4 Newbie Joined: Mar 2014 Posts: 6 Thanks: 0 Re: limit involving 2 sequences How can I actually proof that using what you told me? Do you mean that the limit is actually e?  March 5th, 2014, 11:11 AM #5 Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,821 Thanks: 1047 Math Focus: Elementary mathematics and beyond Re: limit involving 2 sequences Writing out the first few terms of the series, it becomes apparent that a_n/b_n = 1, 1, 1/2, 1/3, 1/4, 1/5, 1/6 . . . We may thus write the limit as $\lim_{n\to\infty}\,n\,\cdot\,$$\frac{1}{n!}$$^{1/n}\,=\,\lim_{n\to\infty}\,$$\frac{n^n}{n!}$$^{1/n}$ Using Stirling's approximation for n!, we have $\lim_{n\to\infty}\,$$\frac{e^n}{\sqrt{2\pi n}}$$^{1/n}\,=\,e$  March 5th, 2014, 07:33 PM #6 Senior Member Joined: Sep 2007 From: USA Posts: 349 Thanks: 67 Math Focus: Calculus Re: limit involving 2 sequences Unfortunately, only testing some terms of $\frac{a_n}{b_n}$ doesn't establish proof that is is true for all values of n. Is it possible to show this limit analytically? My first few attempts tonight were unsuccessful.  March 6th, 2014, 03:12 AM #7 Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,821 Thanks: 1047 Math Focus: Elementary mathematics and beyond Re: limit involving 2 sequences I'm sorry, I should have mentioned that is not a complete proof. My apologies if I haven't been of much assistance.  March 6th, 2014, 06:12 AM #8 Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,821 Thanks: 1047 Math Focus: Elementary mathematics and beyond Re: limit involving 2 sequences I claim that $n\,\cdot\,a_n\,=\,b_n$ Proof: \begin{align*}b_n\,&=\,\[(n\,-\,1)^2\,+\,(n\,-\,1)\,+\,1a_{n-1}\,+\,a_n\,-\,a_{n-1} \\ &=\,(n^2\,-\,n\,+\,1)a_{n-1}\,+\,a_n\,-\,a_{n-1} \\ &=\,n\,\cdot\,a_{n-1}(n\,-\,1)\,+\,a_n \\ &=\,a_n(n\,-\,1)\,+\,a_n\,\,\,(*) \\ &=\,n\,\cdot\,a_n\end{align*} $(*) \\ \text{Claim: }a_n\,=\,n\,\cdot\,a_{n-1} \\ a_1\,=\,2,\,a_2\,=\,4 \/extract_itex] Now it needs to be shown (using the induction hypothesis) that a_{n+1}\,=\,(n\,+\,1)a_n \\ \begin{align*}a_{n+1}\,&=\,a_n\,+\,b_n \\ &=\,a_n\,+\,\[(n\,-\,1)^2\,+\,(n\,-\,1)\,+\,1a_{n-1}\,+\,a_n\,-\,a_{n-1} \\ &=\,n\,\cdot\,a_{n-1}(n\,-\,1)\,+\,2a_n \\ &=\,a_n(n\,-\,1)\,+\,2a_n \\ &=\,(n\,+\,1)a_n\end{align*} Now we may write the limit as $\lim_{n\to\infty}\,n\,\cdot\,$$\frac{1}{n!}$$^{1/n}\,=\,\lim_{n\to\infty}\,$$\frac{n^n}{n!}$$^{1/n}$ Using Stirling's approximation for n!, we have $\lim_{n\to\infty}\,$$\frac{e^n}{\sqrt{2\pi n}}$$^{1/n}\,=\,e$ That's about as analytic as I can get at this time and I can see why there might be problems. My logic is that, for n sufficiently large, the limit may be written as immediately above. Then, $\lim_{n\to\infty}\,$$\frac{e^n}{\sqrt{2\pi n}}$$^{1/n}\,=\,e\lim_{n\to\infty}\,$$\frac{1}{\sqrt{2\pi n}}$$^{1/n}$ So (with L'Hopital's rule), $e\lim_{n\to\infty}\,$$\frac{1}{\sqrt{2\pi n}}$$^{1/n}\,=\,e\lim_{n\to\infty}\exp$$-\frac{1}{2n}\ln(2\pi n)$$\,=\,e\lim_{n\to\infty}\exp$$-\frac{\frac{2\pi}{2\pi n}}{2}$$\,=\,e\,\cdot\,e^0\,=\,e$
 March 6th, 2014, 08:10 AM #9 Newbie   Joined: Mar 2014 Posts: 6 Thanks: 0 Re: limit involving 2 sequences Thank you! After I wrote that $b_{n+1}= (n+1)a_{n+1}$I calculated that limit without using Stirling's approximation. I used d'Alembert's principle and after some arrangements I got :$\lim_{n\to\infty}\frac{(n+1)^{n+1}}{n^n} \frac{a_{n+1}}{b_{n+1}}=>\lim_{n\to\infty}(1+\frac {1}{n})^n=$So this is e. Thanks again!

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